# Means To No End

## When do Means Exist?

Let’s look a bit more closely about the calculation of the arithmetic mean. We noted that m(\lambda i. x_i) + \cdots + m(\lambda i. x_i) = n \cdot m(\lambda i. x_i) and then we noted that multiplying by $n$ is an isomorphism of $\mathbb{R}$. The mean was then given by summing up (taking the aspect of the figure) and then multiplying by the inverse of $n$. We can break this observation up into three parts.

Lemma: If

• $ac^{\ast}$ is a monomorphism, then means are unique,
• $ac^{\ast}$ is a split epimorphism by $s$, then $sa$ is a mean, and
• $ac^{\ast}$ is an isomorphism by $s$, then $sa is the unique mean and it splits$c^{\ast}$$. proof: We prove these claims in turn. • Suppose that $ac^{\ast}$ is mono, and let $m$ and $m'$ be means. Then $ac^{\ast}m = a = ac^{\ast}m'$ and so $m = m'$ by monomorphicity. • Suppose that $ac^{\ast}$ is split epi by $s$. Then $ac^{\ast}s = \textbf{id}$, so that $ac^{\ast}sa = a$. • If $ac^{\ast}$ is iso by $s$, then $sa$ is a mean and since ac^{\ast} is then mono, it is unique. Furthermore, $sac^{\ast} = \textbf{id}$ so that sa splits $c^{\ast}}$. That last property is perhaps useful to single out as a desired characteristic of means. Lacking a better name, I’ll call such a mean normal. Definition: A mean $m$ is normal if $mc^{\ast} = \textbf{id}$. In a category with an epi-mono factorization system, we can factor any aspect $a : X^S \to Q$ into $a = ie$ with $i$ mono and $e$ epi. Since $i$ is mono, the $a$ means correspond to the $e$ means. So, in this case, we can always assume our aspects are epi. ## Perfect Means A perfect mean is a mean which loses no information. In other words, a perfect mean is a mean for the thing itself, and not just some aspect of it. Definition: A perfect mean is a mean for $a = \textbf{id}_{X^S}$. Equivalently, a perfect mean is a section of $\Delta_c$. We can expand the criterion for perfect means a bit. Lemma: If $a$ is mono and $m$ is a mean, then $m$ is a perfect mean. If $m$ is also normal, then $m$ is an inverse of $c^{\ast}$. proof: If $m$ is a mean, then $ac^{\ast}m = a$, so if $a$ is mono, then $c^{\ast}m = \textbf{id}$. For $m$ to be normal is precisely $mc^{\ast} = \textbf{id}$. Intuitively, the existence of a perfect mean should impose strong conditions on the relationship between $S$, $P$, and $X$; if $X$ allows a lot of movement, then this must mean that $S$ and $P$ are very similar, and if $S$ and $P$ are very different, this must mean $X$ is very rigid. We can prove a beginning result to this end. Lemma: Let $S = 2 and$P = 1$. Then$X$admits a perfect mean if and only if it is subterminal. *Proof:* We note that in this case,$c^{\ast} = \Delta$$ is mono. Therefore, it has a section if and only if it is iso. But an object is subterminal if and only if it is isomorphic to its cartesian square.

In other words, if you can perfectly average two things, then you had at most one thing.

## Next Time

In the next post on the topic of means, I hope to look into what happens if you turn the intuition around. The idea of a mean was to express an aspect of a complicated figure in terms of the same aspect of a simpler figure. What if we tried to express an aspect of a simple figure in terms of the same aspect of a more complicated figure?

Written on March 13, 2018